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Can anyone prove this question ?

Let $f$:$\mathbb{R}$$→$$\mathbb{C}$ be continuous function such that $f$$(0)$ $=$ $0$ and that $f'$ be a piecewise continuous function and absolutely integrable on $\mathbb{R}$.

Prove that

$L[f'](s)$ $=$ $sL[f](s)$

where $L[f]$ represents the Laplace transform of f

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2  
Did you try integration by parts? –  mrf May 6 '13 at 21:40
    
what would be my u and v values, if I applied the integration by parts method –  S F May 6 '13 at 21:41
    
Why don't you begin by writing down the definition of the Laplace transform? –  wj32 May 6 '13 at 21:41
    
$f(s)= $ $L(ft)(s)$ $$ \int^inf_0 dx $ –  S F May 6 '13 at 21:44
    
And what does the LHS of the identity you're trying to prove look like? –  wj32 May 6 '13 at 21:47

1 Answer 1

up vote 2 down vote accepted

$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$

Integrating by parts we have,

$L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$

$L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$

If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$

$L\{f'(t)\}=sL\{f(t)\}-f(0)$

Since $f(0)=0$, this reduces to

$L\{f'(t)\}=sL\{f(t)\}$

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thanks, do you know of any online resource that deals with this type of question ? –  S F May 6 '13 at 21:54
    
I think Wiki is sufficient when studying basic Laplace theorem proofs. In the case of this question, wiki goes: en.wikipedia.org/wiki/… –  Maazul May 6 '13 at 22:02

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